I think you mean and not . Very interesting, though.
Define a fibonacci series generated by {a,b} such as,
(The original fibonacci series is generated by {1,1})
( )
Then there are a number of interesting properties:
1)If is the finite continued fraction for
then
Where the appears times.
2)Thus, from here we have that
(Thus, any fibonacci series generated by any two real numbers converges to the divine proportion).
3)The formula for is given by
where is the n-th fibonacci number.
But by Binet's formula we have that,
for the n-th fibonacci number we can find a formula for but it is rather messy and will be omitted. Giving a second method for proving statement 2.
The proof of the second statement is rather neat.
The limit of the ratio of consecutive Fibonacci numbers is already established.
. . . . .
We have:
The ratio is:
Divide top and bottom by
Take the limit:
From [1] amd [2], this becomes:
Multiply top and bottom by
Hello, ThePerfectHacker!
Of course it is . . . I wasn't claiming otherwise.My proof is simpler.
I was demonstrating a straight algebraic approach to the limit
. . for those not familiar with continued fractions.
Please understand: I never ever compete with other posters here.
I may post an alternate approach or even an improvement
. . . but never with a snicker or a sneer, implied or otherwise.